On 03/03/2016 06:40 PM, Mark Dilger wrote:

On Mar 3, 2016, at 8:35 AM, Tomas Vondra <tomas.von...@2ndquadrant.com> wrote:


On 03/03/2016 12:53 PM, Alexander Korotkov wrote:
Hi, Tomas!

I've assigned to review this patch.

I've checked version estimate-num-groups-v2.txt by Mark Dilger.
It applies to head cleanly, passes corrected regression tests.

About correlated/uncorrelated cases. I think now optimizer mostly assumes
all distributions to be independent. I think we should follow this
assumption in this case also until we have fundamentally better option (like
your multivariate statistics).

@@ -3438,9 +3438,9 @@ estimate_num_groups(PlannerInfo *root, List *groupExprs,
double input_rows,
reldistinct = clamp;

-* Multiply by restriction selectivity.
+* Estimate number of distinct values expected in given number of rows.
-reldistinct *= rel->rows / rel->tuples;
+reldistinct *= (1 - powl((reldistinct - 1) / reldistinct, rel->rows));

* Update estimate of total distinct groups.

I think we need way more explanation in comments here (possibly with
external links). For now, it looks like formula which appears from nowhere.

I thought the details (particularly the link to stackexchange, discussing
the formula) would be enough, but let me elaborate.

The current formula

    reldistinct *= rel->rows / rel->tuples;

simply multiplies the ndistinct estimate with selectivity. So when you
select 1% of the table, the estimate says we'll see 1% of ndistinct
values. But clearly that's naive, because for example when you have 10k
distinct values and you select 10M rows, you should expect nearly all of
them in the sample.

The current formula assumes total dependence between the columns.  You can
create tables where this is true, and the formula gives precisely the right
estimate.  I'm not claiming this is common, or that it should be the default
assumption, but it is one end of the spectrum of possible correct

I'm not really sure what you mean by dependence between columns, as both the old and new formula only works with total cardinality and selectivity, and has absolutely no idea about multiple columns.

In case you mean dependence between columns in the GROUP BY and columns used to compute the selectivity (i.e. referenced in WHERE), then perhaps you could say it like that, although I think there's no such explicit assumption.

And that's what the formula does - it gives you the expected number of
distinct values, when selecting 'k' values from 'd' distinct values with

    count(k, d) = d * (1 - ((d-1)/d) ^ k)

It's assuming the distinct values are equally probable (uniform
distribution) and that the probabilities do not change (that's the "with

The new formula assumes total independence between the columns. You can
likewise create tables where this is true, and you did so upthread, and for
those tables it gives precisely the right estimate. This is the other end of
the spectrum of possible correct estimates.

In the absence of multivariate statistics, either because you haven't
committed that work yet, or because the multivariate data the planner needs
hasn't been collected, choosing an estimate exactly in the middle of the
spectrum (whatever that would mean mathematically) would minimize the
maximum possible error in the estimate.

FWIW the current version of multivariate statistics can't really help in this case. Perhaps it will help in the future, but it's way more complicated that it might seem as it requires transferring information from the WHERE clause to the GROUP BY clause.

I have the sense that my point of view is in the minority, because the
expectation is the problems caused by independence assumptions will only be
fixed when multivariate statistics are implemented and available, and so we
should just keep the independence assumptions everywhere until that time. I
 tend to disagree with that, on the grounds that even when that work is
finished, we'll never have complete coverage of every possible set of
columns and what their degree of dependence is.

Well, in a sense you're right that those estimates work best in different situations. The problem with constructing an estimator the way you propose (simply returning an average) is that in both the extreme cases (perfect dependence or independence) one of those estimates is really bad. Moreover the new formula typically produces higher values than the old one,

Consider for example this:

    CREATE TABLE t AS SELECT (10000 * random())::int a,
                             (10000 * random())::int b
                        FROM generate_series(1,1000000) s(i);

and let's see estimates for queries like this:

    SELECT a FROM t WHERE b < $1 GROUP BY a;

Then for different values of $1 we get this:

               |  10 |   50 |  100 |  500 |  1000 |  5000
        actual | 919 | 3829 | 6244 | 9944 | 10001 | 10001
       current |  10 |   50 |  102 |  516 |  1018 |  4996
           new | 973 | 4001 | 6382 | 9897 |  9951 |  9951
       average | 491 | 2025 | 3205 | 5206 |  5484 |  7473

and for a table with perfectly dependent columns:

    CREATE TABLE t AS SELECT mod(i,10000) a,
                             mod(i,10000) b
                       FROM generate_series(1,1000000) s(i);

we get this:

                |  10 |   50 |  100 |  500 |  1000 |  5000
         actual |  10 |   50 |  100 |  500 |  1000 |  5000
        current |  10 |   53 |  105 |  508 |  1016 |  5014
            new | 880 | 4105 | 6472 | 9955 | 10018 | 10018
        average | 445 | 2079 | 3288 | 5231 |  5517 |  7516

So yes, each estimator works great for exactly the opposite cases. But notice that typically, the results of the new formula is much higher than the old one, sometimes by two orders of magnitude (and it shouldn't be difficult to construct examples of much higher differences).

The table also includes the 'average' estimator you propose, but it's rather obvious that the result is always much closer to the new value, simply because

   (small number) + (huge number)

is always much closer to the huge number. We're usually quite happy when the estimates are within the same order of magnitude, so whether it's K or K/2 makes pretty much no difference.


Tomas Vondra                  http://www.2ndQuadrant.com
PostgreSQL Development, 24x7 Support, Remote DBA, Training & Services

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